Chapter 29, Problem 7SP

For the circuit shown in Fig. 29-6, find the current in the 0.96- $\Omega$ resistor and the terminal voltages of the batteries.

Fig. 29-6

To determine

The current in the $0.96\Omega$ resistor and the terminal voltage of the batteries in the circuit given in Fig.29-6.

Answer to Problem 7SP

Solution:

$5.0\text{ A}$, $\text{4}\text{.8 V}$, and $\text{4}\text{.8 V}$

Explanation of Solution

Given data:

Refer to the circuit given in Fig. 29-6.

Formula used:

Kirchhoff’s loop law: The algebraic sum of the potential change in a closed circuit is zero.

In a parallel combination, the voltage remains constant and the current is distributed. In a series combination, the current remains constant and the voltage is distributed.

Write the expression for potential difference from Ohm’s law.

$V=IR$

Here, $R$ is the resistance of the resistor, $V$ is the potential difference across the resistor, and $I$ is the current through the resistor.

Explanation:

Consider the following circuit shown in Figure 1.

Apply loop rule to the loop abcda yields, in volts.

$\begin{array}{c}\left(0.96\right)I_1+\left(0.20\right)\left(I_1-I_2\right)-5.0=0\\ \left(1.16\right)I_1-\left(0.20\right)I_2=5.0\end{array}$ …… (1)

Apply loop rule to loop adefa yields, in volts.

$\begin{array}{c}\left(0.20\right)\left(I_2-I_1\right)+5.0+\left(0.30\right)\left(I_2\right)-6.0=0\\ -\left(0.20\right)I_1+\left(0.50\right)I_2=1.0\end{array}$

Solve further for $I_1$.

$I_2=\frac{1.0+\left(0.20\right)I_1}{\left(0.50\right)}$ …… (2)

Substitute $\frac{1.0\text{ V+}\left(0.20\right)I_1}{\left(0.50\right)}$ for $I_2$ from equation (2) in equation (1)

$\begin{array}{c}\left(1.16\right)I_1-\left(0.20\right)\left(\frac{1.0\text{+}\left(0.20\right)I_1}{\left(0.50\right)}\right)=5.0\\ \left(1.16\right)I_1-0.4-\left(0.08\right)I_1=5.0\\ \left(1.08\right)I_1=5.4\\ I_1=5.0\end{array}$

Thus, this yields that $I_1=5.0\text{}\text{ A}$.

Therefore, the current in the $0.96\Omega$ resistor is $5.0\text{}\text{ A}$.

Calculate the current $I_2$.

Substitute $5.0\text{}$ for $I_1$ in equation (2)

$\begin{array}{c}{I}_2=\frac{1.0+\left(0.20\right)\left(5.0\right)}{\left(0.50\right)}\\ =\frac{2.0}{\left(0.50\right)}\\ =4.0\end{array}$

Thus, this yields that $I_2=4.0\text{}\text{ A}$.

Calculate the current flow in the branch ad in Figure 2.

$\begin{array}{c}{I}_{ad}=I_1-I_2\\ =5.0\text{ A}-4.0\text{ A}\\ \text{=1}\text{.0 A}\end{array}$

The value of $I_{ad}$ is positive. Thus, the direction of the current in $I_{ad}$ is from point d to point a.

Calculate the terminal voltage of the battery, whose emf is ${\epsilon }_1=6.0\text{ V}$.

$V_{T_1}={\epsilon }_1-I_2{r}_1$

Substitute $6.0\text{ V}$ for ${\epsilon }_1$, $4.0\text{}\text{ A}$ for $I_2$, and $0.30\Omega$ for $r_1$

$\begin{array}{c}{V}_{T_1}=6.0\text{ V}-\left(4.0\text{}\text{ A}\right)\left(0.30\Omega \right)\\ =6.0\text{ V}-1.2\text{ V}\\ =4.\text{8 V}\end{array}$

Thus, the terminal voltage of the battery, whose emf ${\epsilon }_1=6.0\text{ V}$ is $4.\text{8 V}$.

Calculate the terminal voltage of the battery, whose emf is ${\epsilon }_2=5.0\text{ V}$.

$V_{T_2}={\epsilon }_2-I_{ad}{r}_2$

Substitute $5.0\text{ V}$ for ${\epsilon }_2$, $1.0\text{}\text{ A}$ for $I_{ad}$, and $0.30\Omega$ for $r_1$

$\begin{array}{c}{V}_{T_2}=6.0\text{ V}-\left(4.0\text{}\text{ A}\right)\left(0.30\Omega \right)\\ =6.0\text{ V}-1.2\text{ V}\\ =4.\text{8 V}\end{array}$

Thus, the terminal voltage of the battery, whose emf ${\epsilon }_2=5.0\text{ V}$ is $4.\text{8 V}$.

Conclusion:

Therefore, the current in the $0.96\Omega$ resistor is $5.0\text{ A}$, the terminal voltage of the battery, whose emf ${\epsilon }_1=6.0\text{ V}$ is $4.\text{8 V}$, and the terminal voltage of the battery, whose emf ${\epsilon }_2=5.0\text{ V}$ is $4.\text{8 V}$.